Answer:
1.0190 x 10⁻⁵ mol
Explanation:
We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).
Molarity = mol/V
V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L
⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L = 1.0190 x 10⁻⁵ mol KIO₃
# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol
The only two that apply is Oxygen and Water. The last one would be Carbohydrates.
The answer would be D.
SiCl4 + 2H2O → SiO2 + 4HCl (balanced equation)
There are 145 centimeters in 1.45 meters.
Answer:
The answer to your question is 1 M
Explanation:
Data
Molarity = ?
mass of CaCl₂ = 222.2 g
Volume = 2 l
Process
1.- Calculate the molar mass of CaCl₂
CaCl₂ = 40 + (35.5 x 2) = 40 + 71 = 111 g
2.- Calculate the moles of CaCl₂
111g of CaCl₂ ---------------- 1 mol
222.2 f of CaCl₂ ---------------- x
x = (222.2 x 1) / 111
x = 222.2 / 111
x = 2 moles
3.- Calculate the Molarity
Molarity = moles / Volume
-Substitution
Molarity = 2/2
-Result
Molarity = 1
